AMC 10 · 2014 · #25

Grade 8 probability

Archetype Branching State Enumeration └ Sub-archetype Markov State Probability

probability-basicrecursive-sequencesystems-of-equationssymmetry-argument convert-to-algebra ↑ Prerequisites: probability-basicsystems-of-equations

📏 Long solution 💡 4 insights

Problem

In a small pond there are eleven lily pads in a row labeled 0 through 10. A frog is sitting on pad 1. When the frog is on pad $N$ , $0<N<10$ , it will jump to pad $N-1$ with probability $\frac{N}{10}$ and to pad $N+1$ with probability $1-\frac{N}{10}$ . Each jump is independent of the previous jumps. If the frog reaches pad 0 it will be eaten by a patiently waiting snake. If the frog reaches pad 10 it will exit the pond, never to return. What is the probability that the frog will escape without being eaten by the snake?

Pick an answer.

(A)

$\frac{32}{79}$

(B)

$\frac{161}{384}$

(C)

$\frac{63}{146}$

(D)

$\frac{7}{16}$

(E)

$\frac{1}{2}$

AMC 10 2014 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

Try it yourself first — the explanation is most useful after you’ve attempted it.

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Toolkit + CCSS Solution

Understand

Restated: A frog starts on pad $1$ of pads numbered $0$ to $10$. From pad $N$ (with $0<N<10$) it hops to $N-1$ with probability $\frac{N}{10}$ and to $N+1$ with probability $1-\frac{N}{10}$. Pad $0$ means eaten, pad $10$ means escape. Find the probability the frog reaches pad $10$ before pad $0$.

Givens: Pads are numbered $0$ through $10$; the frog starts on pad $1$; From pad $N$: down to $N-1$ with probability $\frac{N}{10}$, up to $N+1$ with probability $\frac{10-N}{10}$; Pad $0$ is absorbing (eaten); pad $10$ is absorbing (escape); Jumps are independent

Unknowns: The probability that the frog reaches pad $10$ before pad $0$, starting from pad $1$

Understand

Plan

Primary tool: #4 Introduce a Variable

Secondary: #16 Change Focus / Count the Complement, #7 Identify Subproblems, #11 Work Backwards

There are eleven pads and one wandering frog, so instead of tracking every possible path we name one unknown per pad: $s_N$, the probability of escaping when currently on pad $N$ (Tool #4, Introduce a Variable). One jump then links each $s_N$ to its neighbors, turning the puzzle into a chain of equations. The board is mirror-symmetric about pad $5$, so a change of focus (Tool #16) hands us $s_5=\tfrac12$ for free. Solving the chain one pad at a time (Tool #7, Identify Subproblems) and then back-substituting (Tool #11, Work Backwards) pins down $s_1$.

Execute — Answer: C

#4 Introduce a Variable 6.EE.B.6 Step 1

Name the escape chance per pad

Let $s_N$ be the probability the frog eventually reaches pad $10$ (escapes) given that it is currently on pad $N$.
The two ends are certain: on pad $0$ it is already eaten, and on pad $10$ it has already escaped.
So $s_0=0$ and $s_{10}=1$, and we want $s_1$.

$$s_0 = 0,\qquad s_{10} = 1,\qquad \text{find } s_1$$

💡 Giving every pad its own escape chance turns a random journey into a fixed list of unknown numbers.

#4 Introduce a Variable 7.SP.C.8 Step 2

Link each pad to its neighbors

From pad $N$ the next hop is either down to $N-1$ (probability $\frac{N}{10}$) or up to $N+1$ (probability $\frac{10-N}{10}$); after the hop the frog faces the same problem from the new pad.
By the law of total probability, $s_N=\frac{N}{10}\,s_{N-1}+\frac{10-N}{10}\,s_{N+1}$.
Multiplying by $10$ clears the fractions.

$$10\,s_N = N\,s_{N-1} + (10-N)\,s_{N+1}\qquad(0<N<10)$$

💡 The chance of escaping now is just the weighted average of the escape chances one hop away.

#16 Change Focus / Count the Complement 7.SP.C.7 Step 3

Use symmetry at the middle pad

Pad $5$ sits exactly in the middle.
From it, down and up are equally likely ($\frac{5}{10}=\frac12$ each), and the whole board mirrors: pad $0$ (eaten) and pad $10$ (escape) are the same distance away, with the jump rule reflecting into itself under $N\mapsto 10-N$.
So escaping and being eaten are equally likely from pad $5$.

$$s_5 = \tfrac12$$

💡 A perfectly balanced starting point can only give a fifty-fifty outcome.

#7 Identify Subproblems 8.EE.C.8 Step 4

Climb from the bottom

Plug small $N$ into $10\,s_N = N\,s_{N-1} + (10-N)\,s_{N+1}$ and use $s_0=0$.
For $N=1$: $10\,s_1 = 9\,s_2$, so $s_1=\frac{9}{10}s_2$.
For $N=2$: $10\,s_2 = 2\,s_1 + 8\,s_3$; substituting $s_1$ gives $10\,s_2 = \frac{9}{5}s_2 + 8\,s_3$, so $8\,s_3 = \frac{41}{5}s_2$ and $s_3=\frac{41}{40}s_2$.

$$s_1 = \tfrac{9}{10}s_2,\qquad s_3 = \tfrac{41}{40}s_2$$

💡 Each new equation trades one fresh unknown for the ones already written in terms of $s_2$.

#4 Introduce a Variable 7.EE.B.4 Step 5

Reach the known middle value

For $N=3$: $10\,s_3 = 3\,s_2 + 7\,s_4$.
Using $s_3=\frac{41}{40}s_2$ gives $\frac{41}{4}s_2 = 3\,s_2 + 7\,s_4$, so $s_4=\frac{29}{28}s_2$.
Now $N=4$: $10\,s_4 = 4\,s_3 + 6\,s_5$.
Insert $s_4$, $s_3$, and $s_5=\frac12$: $\frac{145}{14}s_2 = \frac{41}{10}s_2 + 3$.
The $s_2$ terms combine to $\frac{219}{35}s_2 = 3$, giving $s_2=\frac{35}{73}$.

$$\tfrac{219}{35}\,s_2 = 3 \;\Rightarrow\; s_2 = \tfrac{35}{73}$$

💡 Reaching pad $5$'s known $\tfrac12$ closes the chain and finally fixes a real number.

#11 Work Backwards 7.EE.B.4 Step 6

Back-substitute for pad 1

Finally, $s_1=\frac{9}{10}s_2=\frac{9}{10}\cdot\frac{35}{73}=\frac{315}{730}=\frac{63}{146}$.
So the probability the frog escapes starting from pad $1$ is $\frac{63}{146}$, which is choice $\textbf{(C)}$.

$$s_1 = \tfrac{9}{10}\cdot\tfrac{35}{73} = \tfrac{63}{146}\;\Rightarrow\;\textbf{(C)}$$

💡 Once $s_2$ is a number, one multiplication walks the answer back to pad $1$.

[1] #4 6.EE.B.6 Let $s_N$ be the probability the frog eventually reaches pad $10$ (escapes) give

[2] #4 7.SP.C.8 From pad $N$ the next hop is either down to $N-1$ (probability $\frac{N}{10}$) o

[3] #16 7.SP.C.7 Pad $5$ sits exactly in the middle. From it, down and up are equally likely ($\f

[4] #7 8.EE.C.8 Plug small $N$ into $10\,s_N = N\,s_{N-1} + (10-N)\,s_{N+1}$ and use $s_0=0$. Fo

[5] #4 7.EE.B.4 For $N=3$: $10\,s_3 = 3\,s_2 + 7\,s_4$. Using $s_3=\frac{41}{40}s_2$ gives $\fra

[6] #11 7.EE.B.4 Finally, $s_1=\frac{9}{10}s_2=\frac{9}{10}\cdot\frac{35}{73}=\frac{315}{730}=\fr

Review

Reasonableness: $\frac{63}{146}\approx 0.431$ is exactly choice (C), and it is just under $\frac12$. That fits the story: the frog starts on pad $1$, right next to the snake, so escape should be a bit less than even — yet the strong upward push there (up-probability $\frac{9}{10}$) keeps it close. The relation $10\,s_1 = 9\,s_2$ checks out: $9\cdot\frac{35}{73}=\frac{315}{73}$ equals $10\cdot\frac{63}{146}=\frac{630}{146}=\frac{315}{73}$. The symmetric value $s_5=\frac12$ also lands between the endpoints as it must.

Alternative: Write all nine equations for $s_1,\dots,s_9$ and solve the full linear system by elimination without the symmetry shortcut; it returns the same $s_1=\frac{63}{146}$. Alternatively, work with the gaps $d_N=s_N-s_{N-1}$: the recurrence forces $\frac{d_{N+1}}{d_N}=\frac{N}{10-N}$, so each gap is a running product, and summing the gaps from pad $0$ to pad $10$ (which must total $1$) recovers $s_1$ directly.

CCSS standards used (min grade 8)

6.EE.B.6 Use variables to represent numbers and write expressions (Defining $s_N$ as the escape probability from each pad, with $s_0=0$ and $s_{10}=1$.)
7.SP.C.8 Find probabilities of compound events (Writing the one-hop total-probability relation $10\,s_N=N\,s_{N-1}+(10-N)\,s_{N+1}$.)
7.SP.C.7 Develop a probability model and use symmetry (Arguing that the mirror symmetry about pad $5$ forces $s_5=\tfrac12$.)
8.EE.C.8 Solve systems of linear equations (Chaining the pad equations to express $s_1$ and $s_3$ in terms of $s_2$.)
7.EE.B.4 Solve multi-step linear equations (Combining the fractional coefficients to solve $\frac{219}{35}s_2=3$ and back-substituting to $s_1$.)

⭐ Give every pad its own escape chance, use the fair $\tfrac12$ at the middle pad, then walk the equations up from the bottom to pin down pad $1$.

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Name the escape chance per pad

Link each pad to its neighbors

Use symmetry at the middle pad

Climb from the bottom

Reach the known middle value

Back-substitute for pad 1

Review

CCSS standards used (min grade 8)

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